A) a local maxima
B) a local minima
C) neither a local maxima nor a local minima
D) None of the above
Correct Answer: A
Solution :
Let\[f(x)=x\sqrt{1-{{x}^{2}}}\] \[\Rightarrow \] \[f(x)=\frac{1-2{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}=0\] \[\Rightarrow \] \[x=\pm \frac{1}{\sqrt{2}}\] But\[x>0,\]therefore we have \[x=\frac{1}{\sqrt{2}}\] Now, \[f(x)=\frac{\sqrt{1-{{x}^{2}}}(-4x)-(1-2{{x}^{2}})\frac{-x}{\sqrt{1-{{x}^{2}}}}}{(1-{{x}^{2}})}\] \[=\frac{2{{x}^{3}}-3x}{{{(1-{{x}^{2}})}^{3/2}}}\] \[\Rightarrow \] \[f\left( \frac{1}{\sqrt{2}} \right)=-ve,\]maximum. Hence,\[f(x)\] is maximum at\[x=\frac{1}{\sqrt{2}}\].
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